关于高数两个重要极限，是sinX比上X的极限是1(x趋于0）
是等价无穷小的可以化成其他形式，再约分。得到的不一定是1.sinX比上X的极限是1(x趋于0）是由于此时x~sinx
得1。书上例题有很多等价无穷小。看了就懂，记住就会用了（同济6版，58，59页）
绛旓細x鈫0){{[(x^2)/2]^2}^2}/x^4 = 鈥︺備篃鍙埄鐢ㄧ瓑寮 1-cosx = (1/2)[(sinx)^2]锛屽彲寰 g.e. = lim(x鈫0){(1/2)[sin(1-cosx)]^2}/x^4 = 鈥︹ (鍐嶇敤涓娆)= 鈥︺ (鍒╃敤閲嶈鏋侀檺涔嬩竴)
绛旓細鏂规硶涓锛氬ぇ瀛楂樻暟璇炬湰涓殑涓や釜閲嶈鏋侀檺鐨勫叾涓竴涓槸x鈫0鏃讹紝鏈塴im(sin x)/x=1 鎵浠ユ湁lim(sin mx)/(sin nx锛=m/nlim[(sin mx)/(mx)]/[(sin nx/(nx)]=m/n 鏂规硶浜岋細褰搙鈫0鏃讹紝鍒嗗瓙鍒嗘瘝閮解啋0锛屼笉鑳界洿鎺ヤ唬0杩涘幓姹鏋侀檺锛灞炰簬0/0鍨嬶紝鎵浠ュ彲浠ュ埄鐢ㄦ礇蹇呰揪娉曞垯鍋氾紝瀵瑰垎瀛愬垎姣嶅垎鍒眰瀵煎啀...
绛旓細^2 缁撴灉鏄1/2 2.鐢变笁瑙掕瀵煎叕寮忥細tan[蟺/2*(1-X)]=cot蟺X/2鍗虫湁tan(蟺X/2)=1/tan[蟺/2*(1-x)]鏁呭師寮忕瓑浜庯細2/蟺*2/蟺(1-X)*1/tan[蟺/2*(1-x)]缁撴灉鏄2/蟺 3.1/e^2 4.e^2,1,銆楂樻暟銆戝埄鐢涓や釜閲嶈鏋侀檺姹傚嚱鏁版瀬闄 姹備竴涓嬪嚑涓嚱鏁扮殑鏋侀檺,缁撴灉鎴戠煡閬,鎬庝箞鍙樺舰鐨.
绛旓細杩欐槸涓緝涓洪噸瑕鐨鏋侀檺姹傝В锛屼篃姣旇緝鍩烘湰锛屽氨鏄簲鐢╨imx瓒嬭繎浜0锛宻inx~x鐨勭瓑浠蜂唬鎹 1.limx~0鏃讹紝搴旂敤涓婂紡鏈塻in2x~2x,sin5x~5x锛屼笂涓嬪悓鏃剁害鍘粁,寰楀埌绛旀 2/5 2褰搇im n瓒嬭繎浜庢棤绌峰ぇ鏃讹紝x/(2^n)瓒嬭繎浜0锛屾湁sin[x/(2^n)]~x/(2^n),鏈夊師寮忕瓟妗堜负 x锛岄珮鏁杩欐柟闈㈢殑闂浠ュ悗鍙互鎵炬垜锛...
绛旓細杩欐槸涓緝涓洪噸瑕鐨鏋侀檺姹傝В锛屼篃姣旇緝鍩烘湰锛屽氨鏄簲鐢╨imx瓒嬭繎浜0锛宻inx~x鐨勭瓑浠蜂唬鎹 1.limx~0鏃讹紝搴旂敤涓婂紡鏈塻in2x~2x,sin5x~5x锛屼笂涓嬪悓鏃剁害鍘粁,寰楀埌绛旀 2/5 2褰搇im n瓒嬭繎浜庢棤绌峰ぇ鏃讹紝x/(2^n)瓒嬭繎浜0锛屾湁sin[x/(2^n)]~x/(2^n),鏈夊師寮忕瓟妗堜负 x锛岄珮鏁杩欐柟闈㈢殑闂浠ュ悗鍙互鎵...
绛旓細杩涓や釜鏋侀檺鐨閲嶈鎬ф槸鎸囩悊璁轰笂鐨勶細sinx/x鏋侀檺閲嶈鏄鍥犱负鎺ㄥsinx鐨勫鏁版椂瑕佺敤锛屽彟涓涓鏋侀檺鏄e^x瀵兼暟鎺ㄥ鏃剁殑鍩虹銆傚湪鏁板璇剧▼涓瀛︿細濡備綍搴旂敤璇ヤ袱涓瀬闄愭眰鍏朵粬绫讳技鐨勬瀬闄愬笺
绛旓細(b)let y=1/x lim(x->鈭) x.sin(1/x)=lim(y->0) siny/y =1 (d)lim(x->0) 1/(1/x) =0 | sin(1/x) |鈮1 =>lim(x->0) sin(1/x) /(1/x) =0
绛旓細=lim(x->0)[((1/2)/cosx)(sinx/x)(sin(x/2)/(x/2))²] (搴旂敤浣欏鸡鍊嶈鍏紡)=lim(x->0)[(1/2)/cosx]*lim(x->0)[(sinx/x)]*[lim(x->0)(sin(x/2)/(x/2))]²=(1/2)*1*1² (搴旂敤閲嶈鏋侀檺lim(z->0)(sinz/z)=1)=1/2锛沴im(x->1)[(...
绛旓細楂樼瓑鏁板涓鏈夎澶閲嶈鐨鏋侀檺鍏紡锛屽寘鎷絾涓嶉檺浜庝互涓嬪嚑涓細1. 鎸囨暟鍑芥暟鐨勬瀬闄愬叕寮忥細lim(x鈫掆垶) (1 + 1/x)^x = e 2. 鑷劧瀵规暟鍑芥暟鐨勬瀬闄愬叕寮忥細lim(x鈫0) (ln(1 + x))/x = 1 3. 姝ｅ鸡鍑芥暟鐨勬瀬闄愬叕寮忥細lim(x鈫0) (sin x)/x = 1 4. 浣欏鸡鍑芥暟鐨勬瀬闄愬叕寮忥細lim(x鈫0) (1 - ...
绛旓細1銆佸師鏉ョ殑鏋侀檺鏄細lim sin2x/sin5x x鈫0 杩愮敤閲嶈鏋侀檺鍚庯紝鍘熸瀬闄愭垚涓 lim (sin2x/2x)/(sin5x/5x)(5/2) = 2/5 x鈫0 璇存槑锛欰銆(sin2x/2x)(sin5x/5x) 鏄嫾鍑戝嚭鏉ョ殑锛2x璺5x鐨剎鍙互绾﹀垎绾﹀幓锛汢銆佷絾鏄2x鐨2銆5x鐨5锛屾槸鏃犱腑鐢熸湁鐨勶紝蹇呴』涔樹互2/5锛屾墠鑳戒繚鎸佸師棰樼洰涓嶆敼鍙樸傝嫳鏂囦腑鐨...
